Module 6 Teacher Slides

8/11/2019 Module 6 Teacher Slides

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Convection

Heat transfer in the presence of a fluid motion on a solid surface

Various mechanisms at play in the fluid:

- advection physical transport of the fluid- diffusion conduction in the fluid-generation due to fluid friction

But fluid directly in contact with the wall does not move relative to it; hence

direct heat transport to the fluid is by conduction in the fluid only.

T(y)qy y

U T

Ts

u(y)U

TTh

y

Tkq s

y

fconv

0

But depends on the whole fluid motion, and both fluid flow

and heat transfer equations are needed

0

yy

T

T(y)

y TU

Ts


3/40

ConvectionFree or natural convection

(induced by buoyancy

forces)

forced convection (driven

externally)

May occur with

phase change(boiling,

condensation)Convection

Typical values of h (W/m2K)

Free convection: gases: 2 - 25

liquid: 50 - 100

Forced convection: gases: 25 - 250

liquid: 50 - 20,000

Boiling/Condensation: 2500 -100,000

Heat transfer rate q = h( Ts-T )W

h=heat transfer coefficient (W /m2K)

(his not a property. It depends on

geometry ,nature of flow,thermodynamics properties etc.)


4/40

T(y)qy yU T

Ts

u(y)U

Convection rate equation

Main purpose of convective heattransfer analysis is to determine:

flow field

temperature field in fluid

heat transfer coefficient, h

q=heat flux = h(Ts- T)

q = -k(T/ y)y=0

Hence, h = [-k(T/ y)y=0] / (Ts- T)

The expression shows that in order to determine h, we

must first determine the temperature distribution in the

thin fluid layer that coats the wall.


5/40

extremely diverse

several parameters involved (fluid properties, geometry, nature of flow,

phases etc)

systematic approach required

classify flows into certain types, based on certain parameters

identify parameters governing the flow, and group them into meaningful

non-dimensional numbers

need to understand the physics behind each phenomenon

Classes of convective flows:

Common classifications:

A. Based on geometry:

External flow / Internal flowB. Based on driving mechanism

Natural convection / forced convection / mixed convection

C. Based on number of phases

Single phase / multiple phase

D. Based on nature of flowLaminar / turbulent


6/40

How to solve a convection problem ?Solve governing equations along with boundary conditions

Governing equations include

1. conservation of mass2. conservation of momentum

3. conservation of energy

In Conductionproblems, only (3) is needed to be solved.

Hence, only few parametersare involved

In Convection, all the governing equations need to be

solved.

large number of parameterscan be involved


7/40

Nusselt No. Nu = hx / k = (convection heat transfer strength)/(conduction heat transfer strength)

Prandtl No. Pr = / = (momentum diffusivity)/ (thermal diffusivity)

Reynolds No. Re = U x / = (inertia force)/(viscous force)

Viscous force provides the dampening effect for disturbances in the

fluid. If dampening is strong enough laminar flow

Otherwise, instability turbulent flow critical Reynolds number

d

Laminar Turbulent

d

Forced convection: Non-dimensional groupings


8/40

x q

Ts

T

h=f(Fluid, Vel ,Distance,Temp)

Fluid particle adjacent to thesolid surface is at rest

These particles act to retard the

motion of adjoining layers

boundary layer effectMomentum balance:inertia forces, pressure gradient, viscous forces,

body forces

Energy balance: convective flux, diffusive flux, heat generation, energy

storage

FORCED CONVECTION:

external flow (over flat plate)

An internal flow is surrounded by solid boundaries that can restrict the

development of its boundary layer, for example, a pipe flow. An external flow, on

the other hand, are flows over bodies immersed in an unbounded fluid so that the

flow boundary layer can grow freely in one direction. Examples include the flows

over airfoils, ship hulls, turbine blades, etc.


9/40

One of the most important concepts in understanding the external flows is the

boundary layer development. For simplicity, we are going to analyze a boundarylayer flow over a flat plate with no curvature and no external pressure variation.

laminar turbulent

transition

Dye streak

U U UU

Hydrodynamic boundary layer

Boundary layer definitionBoundary layer thickness (d): defined as the distance away from the surfacewhere the local velocity reaches to 99% of the free-stream velocity, that is

u(y=d)=0.99U. Somewhat an easy to understand but arbitrary definition.Boundary layer is usually very thin: d/x usually


10/40

Hydrodynamic and Thermal

boundary layers

As we have seen earlier,the hydrodynamic boundary layer is a region of a

fluid flow, near a solid surface, where the flow patterns are directly

influenced by viscous drag from the surface wall.

0


11/40

Effects of Prandtl number, Pr

ddT

Pr >>1

>> e.g., oils

d, dT

Pr = 1

= e.g., air and gases

have Pr ~ 1

(0.7 - 0.9)

W

W

TTTT

Uu

tosimilar

(Reynolds analogy)

dTd

Pr


12/40

Boundary layer equations (laminar flow)

Simpler than general equations because boundary layer is thin

ddT

U

WT

T

U

x

y

Equations for 2D, laminar, steady boundary layer flow

y

T

yy

Tv

x

Tu

yu

ydxdUU

yuv

xuu

y

v

x

u

:energyofonConservati

:momentum-xofonConservati

0:massofonConservati

Note: for a flat plate, 0hence,constantis dx

dUU


13/40

Exact solutions: Blasius

31

21

31

21

PrRe678.0numberNusseltAverage

PrRe339.0numberNusseltLocal

ReRe

328.11tcoefficiendragAverage

,Re

Re

664.0tcoefficienfrictionSkin

Re

99.4

xknesslayer thicBoundary

0

0

2

21

L

xx

L

L

L

fD

y

wx

x

wf

x

uN

Nu

LUdxCLC

y

uxU

UC

d


14/40

Heat transfer coefficient

Local heat transfer coefficient:

x

k

x

kNuh xxx

31

21

PrRe339.0

Average heat transfer coefficient:

L

k

L

kuNh L

31

21

PrRe678.0

Recall: wallfromrateflowheat, TTAhq wwRecall: wallfromrateflowheat, TTAhq ww

Film temperature, Tfilm

For heated or cooled surfaces, the thermophysical properties within

the boundary layer should be selected based on the average

temperature of the wall and the free stream;

TTT wfilm 21


15/40

Heat transfer coefficient

U

x

HydrodynamicBoundary Layer, d

Convection

Coefficient, h.

Thermal Boundary

Layer, dt

Laminar Region Turbulent Region

Laminar and turbulent b.l.


16/40

Turbulent boundary layer

etc.:wayusualintcoefficienferheat transCalculate*

Re664.0Re036.0PrPrRe036.0

PrRe029.0

Re328.1Re072.0Re

1Re072.0

)105(ReRe059.0

:dataalexperimentonbasednscorrelatiousemainlywillWe*

solve.todifficultmore

infinitelybutones,laminarsimilar toareequationsb.l.Turbulent*

)105plate,flatoverflowFor(Re

.turbulenteventuallyandnaltransitiobecomesflowthe),ReRe(

numberReynoldsofvaluecriticalaBeyondwith x.increasesRe*

5.08.08.0

8.0

5.08.0

52.0

5

31

31

31

x

kNu

h

uN

Nu

C

C

xxU

xcxcL

xx

xcxcL

LD

xxf

cc

xc

xcx

x


17/40

d x( )

x

0 0.5 10

0.5

1

Boundary layer growth: d xInitial growth is fast

Growth rate dd/dx 1/x,decreasing downstream.

w x( )

x

0 0.50

5

10

Wall shear stress: w1/x

As the boundary layer grows, thewall shear stress decreases as the

velocity gradient at the wall becomes

less steep.

Laminar Boundary Layer Development


18/40

Determine the boundary layer thickness, the wall shear stress of a laminar water flow

over a flat plate. The free stream velocity is 1 m/s, the kinematic viscosity of the wateris 10-6m2/s. The density of the water is 1,000 kg/m3. The transition Reynolds number

Re=Ux/=5105. Determine the distance downstream of the leading edge when theboundary transitions to turbulent. Determine the total frictional drag produced by the

laminar and turbulent portions of the plate which is 1 m long. If the free stream and

plate temperatures are 100 C and 25 C, respectively, determine the heat transfer ratefrom the plate.

3

2

w

( ) 5 5 10 ( ).

Therefore, for a 1m long plate, the boundary layer grows by 0.005(m),

or 5 mm, a very thin layer.

0.332 0.0105The wall shear stress, 0.332 ( )

Re

The transition Reyn

x

xx x m

U

U UU Pa

x x

d

5Uolds number: Re 5 10 , 0.5( )tr trx

x m

Example


19/40

2

D

0 0

f

2

The total frictional drag is equal to the integration of the wall shear stress:

0.664F (1) 0.332 0.939( )

Re

Define skin friction coefficient: C

0.664for a

1 Re2

tr tr

tr

x x

w

x

wf

x

U Udx U dx N

x

CU

laminar boundary layer.

Example (cont..)


20/40

DUD

Re

Perimeter

Area4hD

D

kNh

k

DhN

AU

C uunormal

D

;;forceDrag

2

21

Forced convection over exterior bodies

Much more complicated.

Some boundary layer may exist, but it is likelyto be curved and Uwill not be constant.Boundary layer may also separate from the

wall.

Correlations based on experimental data canbe used for flow and heat transfer calculations

Reynolds number should now be based on a

characteristic diameter.

If body is not circular, the equivalentdiameter Dh is used


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0.70.0810-102

0.60.26102-10

0.50.5110-40

0.40.75401

mCRe

Pr

PrReuN

65

53

3

D

25.

62.

s

m

DC

Flow over circular cylinders

All properties at free stream

temperature, Prsat cylinder

surface temperature


22/40

Thermal conditions

Laminar or turbulententrance flow and fully developed thermal condition

For laminar flows the thermal entrance length is a function of theReynolds number and the Prandtl number: xfd,t/D 0.05ReDPr,where the Prandtl number is defined as Pr = /and is the thermaldiffusitivity.

For turbulent flow, xfd,t10D.

FORCED CONVECTION: Internal flow

Thermal entrance region, xfd,t

e.g. pipe flow


23/40

Thermal Conditions

For a fully developed pipe flow, the convection

coefficient does not vary along the pipe length.

(provided all thermal and flow properties are constant)

x

h(x)

xfd,t

constant

Newtons law of cooling: qS = hA(TS-Tm)

Question: since the temperature inside a pipe flow does not remain

constant, we should use a mean temperature Tm, which is defines

as follows:


24/40

Energy Transfer

Consider the total thermal energy carried by the fluid as

(mass flux) (internal energy)vA

VC TdA Now imagine this same amount of energy is carried by a body

of fluid with the same mass flow rate but at a uniform mean

temperature Tm. Therefore Tmcan be defined as

v

Am

v

VC TdA

TmC

Consider Tmas the reference temperature of the fluid so that the

total heat transfer between the pipe and the fluid is governed by theNewtons cooling law as: qs=h(Ts-Tm), where h is the local

convection coefficient, and Tsis the local surface temperature.

Note: usually Tmis not a constant and it varies along the pipe

depending on the condition of the heat transfer.


25/40

Energy Balance

Example: We would like to designa solar water heater that can heat up the watertemperature from 20 C to 50 C at a water flow rate of 0.15 kg/s. The water is

flowing through a 0.05 m diameter pipe and is receiving a net solar radiation

flux of 200 W/m of pipe length. Determine the total pipe length required to

achieve the goal.


26/40

Example (cont.)

Questions: (1) How to determine the heat transfer coefficient, h?

There are a total of six parameters involved in this problem:h, V, D, , kf,cp. The temperature dependence of properties is implicit and is only

through the variation of thermal properties. Density is included in the

kinematic viscosity, /. According to the Buckingham theorem, it ispossible for us to reduce the number of parameters by three. Therefore, theconvection coefficient relationship can be reduced to a function of only

three variables:

Nu=hD/kf, Nusselt number, Re=VD/, Reynolds number, andPr=/, Prandtl number.

This conclusion is consistent with empirical observation, that is

Nu=f(Re, Pr). If we can determine the Reynolds and the Prandtl numbers,

we can find the Nusselt number and hence, the heat transfer coefficient, h.


27/40

Convection Correlations

ln(Nu)

ln(Re)

slope m

Fixed Pr

ln(Nu)

ln(Pr)

slope n

Fixed Re

D s

D s

Laminar, fully developed circular pipe flow:

Nu 4.36, when q " constant, (page 543, ch. 10-6, ITHT)

Nu 3.66, when T constant, (page 543, ch. 10-6, ITHT)

Note: t

f

hD

k

mhe therma conductivity should be calculated at T .

Fully developed, turbulent pipe flow: Nu f(Re, Pr),

Nu can be related to Re & Pr experimentally, as shown.

E i i l C l i


28/40

Empirical Correlations

4/5

D

s m s m

D

Dittus-Boelter equation: Nu 0.023Re Pr , (eq 10-76, p 546, ITHT)

where n 0.4 for heating (T T ), n 0.3 for cooling (T T ).

The range of validity: 0.7 Pr 160, Re 10,000, / 10.

n

L D

Note: This equation can be used only for moderate temperature difference with all

the properties evaluated at Tm.

Other more accurate correlation equations can be found in other references.

Caution: The ranges of application for these correlations can be quite different.

D1/ 2 2 / 3

D

( / 8)(Re 1000) Pr Nu (from other reference)

1 12.7( / 8) (Pr 1)

It is valid for 0.5 Pr 2000 and 3000 Re 5 10

Df

f

6

m

.

All properties are calculated at T .

E l ( t )


29/40

Example (cont.)

In our example, we need to first calculate the Reynolds number: water at 35C,Cp=4.18(kJ/kg.K), =7x10-4(N.s/m2), kf=0.626 (W/m.K), Pr=4.8.

4

D 1/ 2 2 / 3

4 4(0.15)Re 5460

(0.05)(7 10 )

Re 4000, it is turbulent pipe flow.

Use the Gnielinski correlation, from the Moody chart, f 0.036, Pr 4.8

( / 8)(Re 1000) Pr (0.Nu

1 12.7( / 8) (Pr 1)

D

m DVD mA

D

f

f

1/ 2 2 / 3

2

036 / 8)(5460 1000)(4.8)37.4

1 12.7(0.036 / 8) (4.8 1)

0.626(37.4) 469( / . )

0.05

f

D

kh Nu W m K

D

l


30/40

Energy Balance

Question (2): How can we determine the required pipe length?Use energy balance concept: (energy storage) = (energy in) minus (energy out).

energy in = energy received during a steady state operation (assume no loss)

'( ) ( ),

( ) (0.15)(4180)(50 20)94( )

' 200

P out in

P in out

q L mC T T

mC T T L m

q

q=q/L

Tin Tout


31/40

Temperature Distribution

s s

s

From local Newton's cooling law:q hA(T ) ' ( )(T ( ) ( ))

' 200( ) ( ) 20 0.319 22.7 0.319 ( )

(0.05)(469)

At the end of the pipe, T ( 94) 52.7( )

m m

s m

T q x h D x x T x

qT x T x x x C

Dh

x C

Question (3): Can we determine the water temperature variation along the pipe?

Recognize the fact that the energy balance equation is valid for

any pipe length x:

'( ) ( ( ) )

' 200( ) 20 20 0.319(0.15)(4180)

It is a linear distribution along the pipe

P in

in

P

q x mC T x T

qT x T x x xmC

Question (4): How about the surface temperature distribution?


32/40

Temperature variation for constant heat flux

Note: These distributions are valid only in the fully developed region. In the

entrance region, the convection condition should be different. In general, the

entrance length x/D10 for a turbulent pipe flow and is usually negligible as

compared to the total pipe length.

T m x( )

T s x( )

x

0 20 40 60 80 10020

30

40

50

60

Constant temperature

difference due to the

constant heat flux.

I t l Fl C ti


33/40

Internal Flow Convection

-constant surface temperature case

Another commonly encountered internal convection condition is when the

surface temperature of the pipe is a constant. The temperature distribution in

this case is drastically different from that of a constant heat flux case. Consider

the following pipe flow configuration:

Tm,

i

Tm,oConstant Ts

Tm Tm+dTm

qs=hA(Ts-Tm)

p

p

s

p s

Energy change mC [( ) ]

mC

Energy in hA(T )

Energy change energy in

mC hA(T )

m m m

m

m

m m

T dT T

dT

T

dT T

dx


34/40

Temperature distribution

p s

s m

m

mC hA(T ),

Note: q hA(T ) is valid locally only, since T is not a constant

, where A Pdx, and P is the perimeter of the pipe(T )

Integrate from the inlet to a diatance x downstr

m m

m

m

s P

dT T

T

dT hA

T mC

,

,

( )

0 0m

( )

m

0

eam:

(T )

ln(T ) | , where L is the total pipe length

and h is the averaged convection coefficient of the pipe between 0 & x.

1, or

m

m i

m

m i

T x x xm

Ts P P

T x

s T

P

x

dT hP Pdx hdx

T mC mC

PhT x

mC

h hdx hdxx

0x

hx

T di ib i


35/40

Temperature distribution

,

( )exp( ), for constant surface temperaturem s

m i s P

T x T Phx

T T mC

T x( )

x

Tm(x)

Constant surface temperature

Ts

The difference between the averaged fluid temperature and the surface

temperature decreases exponentially further downstream along the pipe.

L M T Diff


36/40

Log-Mean Temperature Difference

,

,

, , , ,

For the entire pipe:

( )exp( ) or

ln( )

( ) (( ) ( ))

( )

ln( )

where is calln( )

m o s o sP

om i s i P

i

P m o m i P s m i s m o

o iP i o s s lm

o

i

o i

lmo

i

T T T hAh PL mCTT T T mC

T

q mC T T mC T T T T

T TmC T T hA hA T

T

T

T T

T T

T

led the log mean temperature difference.

This relation is valid for the entire pipe.

F C i


37/40

Free Convection

A free convection flow field is a self-sustained flow driven by the

presence of a temperature gradient. (As opposed to a forcedconvection flow where external means are used to provide the flow.)

As a result of the temperature difference, the density field is not

uniform also. Buoyancy will induce a flow current due to the

gravitational field and the variation in the density field. In general,a free convection heat transfer is usually much smaller compared to

a forced convection heat transfer. It is therefore important only

when there is no external flow exists.

hot

cold

T T

Flow is unstable and a circulatory

pattern will be induced.

B i D fi iti


38/40

Basic Definitions

Buoyancy effect:

Warm,

Surrounding fluid, cold,

Hot plateNet force=(- gV

The density difference is due to the temperature difference and it can be

characterized by ther volumetric thermal expansion coefficient, b:1 1 1

( )PT T T T

T

b

b

G h f N b d R l i h N b


39/40

Grashof Number and Rayleigh Number

Define Grashof number, Gr, as the ratio between the buoyancy force and the

viscous force: 33

2 2

( )Sg T T Lg TLGr bb

Grashof number replaces the Reynolds number in the convection correlation

equation. In free convection, buoyancy driven flow sometimes dominates the

flow inertia, therefore, the Nusselt number is a function of the Grashof numberand the Prandtle number alone. Nu=f(Gr, Pr). Reynolds number will be

important if there is an external flow. (combined forced and free convection.

In many instances, it is better to combine the Grashof number and the

Prandtle number to define a new parameter, the Rayleigh number, Ra=GrPr.The most important use of the Rayleigh number is to characterize the laminar

to turbulence transition of a free convection boundary layer flow. For

example, when Ra>109, the vertical free convection boundary layer flow over

a flat plate becomes turbulent.

Example


40/40

Example

Determine the rate of heat loss from a heated pipe as a result of

natural (free) convection.

Ts=100C

T=0C D=0.1 m

Film temperature( Tf): averaged boundary layer temperature Tf=1/2(Ts+T )=50 C.

kf=0.03 W/m.K, Pr=0.7, =210-5m2/s, b=1/Tf=1/(273+50)=0.0031(1/K)

3 36

2 5 2

1/ 62

9 /16 8 /27

2

( ) (9.8)(0.0031)(100 0)(0.1)Pr (0.7) 7.6 10 .

(2 10 )

0.387

{0.6 } 26.0 (equation 11.15 in Table 11.1)[1 (0.559 / Pr) ]

0.03(26) 7.8( / )

0.1

( ) (7.8)( )(

S

D

f

D

S

g T T LRa

Ra

Nu

kh Nu W m K

D

q hA T T

b

0.1)(1)(100 0) 244.9( )

C b i ifi if h i l

W

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